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Last updated at Jan. 29, 2020 by Teachoo

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Example 10 Find the term independent of x in the expansion of (3/2 ๐ฅ^2 " โ " 1/3๐ฅ)^6,x > 0. Calculating general term We know that general term of expansion (a + b)n is Tr + 1 = nCr (a)nโr.(b)n For general term of expansion (3/2 ๐ฅ^2 " โ " 1/3๐ฅ)^6 Putting n = 6 , a = 3/2 ๐ฅ^2 , b = "โ" 1/3๐ฅ Tr + 1 = 6Cr (๐/๐ ๐^๐ )^(๐ โ ๐) ((โ๐)/๐๐)^๐ = 6Cr (3/2)^(6 โ ๐) (๐ฅ^2 ")" ^((6 โ ๐)) ((โ1)/3 "ร" 1/๐ฅ)^๐ = 6Cr (3/2)^(6 โ ๐) (๐ฅ")" ^(2(6 โ ๐)) ((โ1)/3)^๐ (1/๐ฅ)^๐ = 6Cr (3/2)^(6 โ ๐) (๐ฅ")" ^(12 โ 2๐) ((โ1)/3)^๐ (๐ฅ)^(โ๐) = 6Cr (3/2)^(6 โ ๐) ((โ1)/3)^๐ (๐ฅ")" ^(12 โ 2๐) (๐ฅ)^(โ๐) = 6Cr (3/2)^(6 โ ๐) ((โ1)/3)^๐ (๐ฅ")" ^(12 โ 2๐ โ ๐) = 6Cr (3/2)^(6 โ ๐) ((โ1)/3)^๐ (๐ฅ")" ^(12 โ3๐) We need to find the term independent of x So, power of x is 0 ๐ฅ^(12 โ 3๐) = x0 Comparing powers 12 โ 3r = 0 12 = 3r 12/3 = r 4 = r r = 4 Putting r = 4 in (1) T4+1 = 6C4 (3/2)^(6 โ 4) ((โ1)/3)^4 (๐ฅ")" ^(12 โ3(4)) T5 = 6C4 (3/2)^2 (1/3^4 ) (๐ฅ")" ^(12 โ12) = 6C4 (3^2/2^2 )(1/3^4 ) (๐ฅ")" ^0 = 6C4 (1/2^2 )(3^2/3^4 ) (1) = 6C4 (1/2^2 )(1/3^2 ) = 6!/4!(6 โ 4)! (1/4)(1/9) = 6!/4!(2)! (1/4)(1/9) = (6(5)(4)!)/4!(2)! (1/4)(1/9) = (5 )/12 Hence, the term which is independent of x is 5th term = T5 = ๐/๐๐

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.